$f(x, y) = (xy)^5$ $\dfrac{\partial^3 f}{\partial y^3} = $
Solution: Taking a third order partial derivative is like taking a regular third order derivative. We take the partial derivative once, then we take another partial derivative, and then we take one more. $\dfrac{\partial^3 f}{\partial y^3} = \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial y} \right] \right]$ Let's differentiate! $\begin{aligned} \dfrac{\partial^3 f}{\partial y^3} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial y} \left[ (xy)^5 \right] \right] \right] \\ \\ &= x^5 \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial y} \left[ 5y^4 \right] \right] \\ \\ &= x^5 \dfrac{\partial}{\partial y} \left[ 20y^3 \right] \\ \\ &= 60 x^5 y^2 \end{aligned}$ Therefore, $\dfrac{\partial^3 f}{\partial y^3} = 60 x^5 y^2$.